# Core 4

- You can find the formula booklet here and past papers from 2005 to present here. To convert your scores to UMS marks, my online converter tool and Android app are linked from the homepage of jameswalker.net.

To use the flashcards, you'll probably need a paper and pen. You shouldn't need the formula booklet

**All formulae which are in the formula booklet are coloured in blue**$$

# Algebra

**a1** Binomial expansion of $(1+x{)}^{n}$ :

- If $x$ is a positive integer, $(1+x{)}^{n}$ can be expanded with the method from C1
- For all values of $x$ where $|x|<1$, $(1+x{)}^{n}=1+nx+\frac{n(n-1)}{2!}{x}^{2}+\frac{n(n-1)(n-2)}{3!}{x}^{3}+...$ to infinity
- Although this can require infinite terms to be accurate, we can approximate the expansion by giving the first few terms
*Note for FP2 students - you need to be able to use this formula and recognise that it's a Maclaurin series (it can be derived using the chain rule)*

## Example: find the first three terms of $(1+x{)}^{-2}$:

- $n=-2$
- Using the formula, the first three terms are $1-2x+\frac{-2(-2-1)}{2!}{x}^{2}$
- This simplifies to $1-2x+3{x}^{2}$

## Example: find the first three terms of ${}^{1}{/}_{1+2x}$:

- This is the same as $(1+2x{)}^{-1}$
- So $n=-1$
- We still have $2x$ instead of $x$. So we replace all instances of $x$ in the expansion formula with $(2x)$ (brackets are important!)
- Using the formula, the first three terms are $1-(2x)+\frac{-1(-1-1)}{2!}(2x{)}^{2}$
- This simplifies to $1-2x+4{x}^{2}$
- Usually, the range of $x$ would be $|x|<1$. However, we are now using $2x$, so the range this time would be $|2x|<1$, which simplifies to $|x|<\frac{1}{2}$

**a2** Binomial expansion of $(a+x{)}^{n}$ :

- To get $(a+x{)}^{n}$ into a form which can be expanded, we need to take $a$ outside of the bracket with the following steps:

$(a+x{)}^{n}$

$(a(1+\frac{x}{a}){)}^{n}$

${a}^{n}(1+\frac{x}{a}{)}^{n}$ - Now we have the first term of the bracket equal to 1, so we can expand

## Example: find the first three terms of $(3+x{)}^{-1}$:

- With the above rearrangement method, this rearranges to ${3}^{-1}(1+\frac{x}{3}{)}^{-1}$
- We can now expand the inside of the bracket to $1-\frac{x}{3}+\frac{-1(-1-1)}{2!}{\frac{x}{3}}^{2}$
- This simplifies to $1-\frac{x}{3}+\frac{{x}^{2}}{9}$
- Now we need to multiply this by the number to the left of the bracket, ${3}^{-1}$
- Therefore, our final expansion is $\frac{1}{3}-\frac{x}{9}+\frac{{x}^{2}}{27}$
- The range of $x$ is $|\frac{x}{3}|<1$, which simplifies to $|x|<3$

**a3** Simplifying rational fractions:

- a3 is just GCSE fractions so I won't cover it here. It's in pages 166 - 168 of the textbook
- This is: factorising, cancelling and algebraic fractions

**a4** Solving equations involving algebraic fractions:

- Again, a4 is GCSE level. It's in pages 169 - 172 in the textbook

**a5** Partial fractions:

- Sometimes we need to write a single fraction as the sum of two separate fractions - e.g. for integration and binomial expansion

## General method for denominators of the form $(ax+b)(cx+d)$:

- As an example, I'll use the fraction $\frac{5}{(x-2)(x+3)}$
- We can write this as $\frac{A}{x-2}$ $-$ $\frac{B}{x+3}$. Notice that each fraction has the same denominator as one of the brackets in the original fraction. $A$ and $B$ are constants which we'll need to calculate
- To add fractions, they need to have the same denominator. So we can cross-multiply the two fractions to get $\frac{A(x+3)}{(x-2)(x+3)}$ $+$ $\frac{B(x-2)}{(x-2)(x+3)}$
- Now we are able to multiply these two fractions to get $\frac{A(x+3)+B(x-2)}{(x-2)(x+3)}$
- Notice that this has the same denominator as our original fraction. It is also equal, as we haven't changed it in any of the steps, it's only been rearranged. Therefore, we can set the two equal: $\frac{A(x+3)+B(x-2)}{(x-2)(x+3)}$ $=$ $\frac{5}{(x-2)(x+3)}$
- Since the denominators are the same, we can compare the numerators: $A(x+3)+B(x-2)=5$
- This expands to $Ax+3A+Bx-2B=5$
- We can factorise this to $x(A+B)+(3A-2B)=x(0)+(5)$. This allows us to set up two simultaneous equations:

$A+B=0$

$3A-2B=5$ - Now use a GCSE method to solve these equations. I'll multiply the first by 2 and add them together to $5A=5$. Therefore, $A=1$. Looking at the first equation, we can see that $B=-1$
- We wrote the original fraction as two partial fractions in terms of $A$ and $B$ in the second bullet point. Now we can substitute in the values to get $\frac{1}{x-2}$ $+$ $\frac{-1}{x+3}$ which simplifies to $\frac{1}{x-2}-\frac{1}{x+3}$

## For denominators of the form $(ax+b)(c{x}^{2}+d)$:

- Instead of $A$ and $B$ for the constants, use $A$ and $Bx+C$
- Use the above method. You should get three equations to solve. To solve them, rearrange one of them so that you can substitute it into another of the equations. This will leave you with two equations, each with two terms
- For example, if you get the equations

$A+B=0$

$-B+C=2$

$4A-C=3$,

you can rearrange the first to $-B=A$ and substitute this into the second equation. This would leave you with

$A+C=2$

$4A-C=3$.

These can be solved by adding the two equations together

## For denominators of the form $(ax+b)(cx+d{)}^{2}$:

- Write this as three partial fractions with numerators $A$, $B$ and $C$
- The denominators should be $ax+b$, $cx+d$ and $(cx+d{)}^{2}$
- Once you have combined the three partial fractions, you can divide both sides by $(cx+d)$
- From here, the method is the same as usual

**a6** Using partial fractions with the binomial expansion:

- For binomial expansion, you need the form $(1\pm x{)}^{n}$
- So you'll need to expand each partial fraction separately and sum them
- If the numerator of a partial fraction is not $1$, move it out of the fraction. For example, if you have the partial fraction $\frac{5}{x+1}$, you can write this as $5\times \frac{1}{x+1}$ and then $5(1+x{)}^{-1}$
- If the denominator contains a constant that is not $\pm x$, you can use the same procedure. For example, if you have the partial fraction $\frac{1}{x+2}$, you can factorise the denominator to $2(1+\frac{x}{2})$ to make $\frac{1}{2(1+\frac{x}{2})}$. Now move the $\frac{1}{2}$ outside the fraction; $\frac{1}{2}(1+\frac{x}{2}{)}^{-1}$

# Trigonometry

**t1** Reciprocal trigonometrical functions:

- $\mathrm{cosec}\theta =$ $\frac{1}{\mathrm{sin}\theta}$
- $\mathrm{sec}\theta =$ $\frac{1}{\mathrm{cos}\theta}$
- $\mathrm{cot}\theta =$ $\frac{1}{\mathrm{tan}\theta}$
- You can remember these by looking at the third letter of the reciprocal function name - the third letter of $\mathrm{cosec}$ is $s$ (first letter of $\mathrm{sin}$), the third letter of $\mathrm{sec}$ is $c$ (first letter of $\mathrm{cos}$), etc
- These functions are undefined when the denominator would be $0$. For example, $\mathrm{cosec}\theta $ is undefined when $\mathrm{sin}\theta =0$: ${0}^{\circ},\text{}{180}^{\circ},\text{}{360}^{\circ},\text{}...$

**t2** Reciprocal trigonometrical function graphs:

- In the below graphs, the red line is the original trigonometrical function and the blue line is the reciprocal function
- $y=\mathrm{cosec}x$ has vertical asymptotes every ${180}^{\circ}$, starting at ${0}^{\circ}$
- $y=\mathrm{sec}x$ has vertical asymptotes every ${180}^{\circ}$, starting at ${90}^{\circ}$
- $y=\mathrm{cot}x$ has vertical asymptotes every ${180}^{\circ}$, starting at ${0}^{\circ}$

**t3** Reciprocal trigonometrical function relationships:

- The following can sometimes be used for simplification:

${\mathrm{tan}}^{2}\theta +1\equiv {\mathrm{sec}}^{2}\theta $

${\mathrm{cot}}^{2}\theta +1\equiv {\mathrm{cosec}}^{2}\theta $ - You can derive them by starting with ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \equiv 1$ (from C2) and dividing both sides by ${\mathrm{sin}}^{2}\theta $ or ${\mathrm{cos}}^{2}\theta $

**t4** Compound angle formulae:

- Compound angle formulae are used when you have an angle containing two known angles. We call these angles $\theta $ and $\varphi $. They are given in the formula booklet in the following form:

$\mathrm{sin}(\theta \pm \varphi )=\mathrm{sin}\theta \mathrm{cos}\varphi \pm \mathrm{cos}\theta \mathrm{sin}\varphi $

The two $\pm $s should be equal (i.e. both $+$ or both $-$)

$\mathrm{cos}(\theta \pm \varphi )=\mathrm{cos}\theta \mathrm{cos}\varphi \mp \mathrm{sin}\theta \mathrm{sin}\varphi $

The $\pm $ and $\mp $ should have opposite signs in this formula

$\mathrm{tan}(\theta \pm \varphi )=$ $\frac{\mathrm{tan}\theta \pm \mathrm{tan}\varphi}{1\mp \mathrm{tan}\theta \mathrm{tan}\varphi}$

Again, the $\pm $ and $\mp $ should be opposite signs, and both $\pm $s should have the same sign

**t5** Double angle formulae:

- Double angle formulae are not given in the formula booklet. However, they are easily derived from identities that are:
- $\mathrm{sin}2\theta =2\mathrm{sin}\theta \mathrm{cos}\theta $

- This can be derived by using the compound angle formula for $\mathrm{sin}(\theta +\theta )$ - $\mathrm{cos}2\theta ={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta $

- This can be derived using the compound angle formula for $\mathrm{cos}(\theta +\theta )$

- By using the identity ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$, you can use this to get the following:

- $\mathrm{cos}2\theta =1-2{\mathrm{sin}}^{2}\theta $

- $\mathrm{cos}2\theta =2{\mathrm{cos}}^{2}\theta -1$ - $\mathrm{tan}2\theta =$ $\frac{2\mathrm{tan}\theta}{1-{\mathrm{tan}}^{2}\theta}$

- This can be derived from the compound angle formuala for $\mathrm{tan}(\theta +\theta )$

**t6** Solving trigonometrical equations:

- To solve a trigonometrical equation, you often need to simplify and rearrange
- You can use the identities above to do this

**t7** Writing in the forms $R\mathrm{sin}(\theta \pm \alpha )$ and $R\mathrm{cos}(\theta \pm \alpha )$:

- Any expression of the form $a\mathrm{cos}\theta +b\mathrm{sin}\theta $ can be written in this form

## Steps:

- Here we'll use the example $3\mathrm{cos}\theta +4\mathrm{sin}\theta $. We'll be writing it in the form $R\mathrm{cos}(\theta -\alpha )$
- Write what you have and what you want to get:

$3\mathrm{cos}\theta +4\mathrm{sin}\theta =R\mathrm{cos}(\theta -\alpha )$ - This can be expanded using a compound angle formula to

$3\mathrm{cos}\theta +4\mathrm{sin}\theta =R\mathrm{cos}\theta \mathrm{cos}\alpha +R\mathrm{sin}\theta \mathrm{sin}\alpha $ - You'll notice that if $R\mathrm{cos}\alpha =3$, then the first term on the left would be equal to the first term on the right
- You'll also find that the same applies to the second term if $R\mathrm{sin}\alpha =4$
- Now you'll need to solve the two above equations to find $R$ and $\alpha $
- If you divide the $\mathrm{sin}$ equation by the $\mathrm{cos}$ one, you'll get

$\frac{R\mathrm{sin}\alpha}{R\mathrm{cos}\alpha}$ $=$ $\frac{4}{3}$

This simplifies to $\mathrm{tan}\alpha =\frac{4}{3}$

Therefore, $\alpha =\mathrm{arctan}\frac{4}{3}={53.1}^{\circ}$ - A quick way to find $R$ is to find the square root of the sum of the two answers of the simultaneous equations squared, i.e. $R=\sqrt{{3}^{2}+{4}^{2}}$ for our example. This method is derived by squaring the two simultaneous equations and using the identity ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$

## Summary of steps:

- The following is the quickest method which would gain you all method marks in the exam
- Expand the form which the question asks for (e.g. $R\mathrm{cos}(\theta -\alpha )$) with a compound angle formula
- Write the two simultaneous equations
- Square the two integer answers to these equations, sum them and take the square root to find $R$
- Find $\mathrm{arctan}$ of the $\mathrm{sin}$ term divided by the $\mathrm{cos}$ term to find $\alpha $. Use an exact answer, or $3$ significant figures
- Substitute your values into $R$ and $\alpha $ in the form the question asks for

## Minimum and maximum:

- The minimum and maximum are $\pm R$

## Sketching the function:

- $y=R\mathrm{cos}(\theta -\alpha )$ is a graph of $y=\mathrm{cos}\theta $ stretched with scale factor $R$ parallel to the $y$ axis and translated $\alpha $ in the positive direction (the opposite sign to the sign inside the brackets)
- Use the graph to solve equations in this form

# Parametric equations

**g1** Introduction:

- A
**parametric equation**is a set of equations with two variables (like in all equations) plus at least one additional variable (**parameter**) - For example, you could have

$x=2\mathrm{cos}\theta $

$y=2\mathrm{sin}\theta $

where $\theta $ is the*parameter*

**g2** Converting parametric equations to cartesian equations:

- A
**cartesian equation**relates two points on a $x$ and $y$ coordinate system. Examples include $y=2x$ and ${x}^{2}+{y}^{2}=4$ - Two methods are shown below (only one of them will usually work well)

## Method 1:

- You'll need to rearrange one equation so that you can substitute it into another, eliminating the parameter
- Take $x={t}^{2}$ and $y=2t$

- Pick an equation and rearrange it in terms of the parameter ($t$): $t=\frac{y}{2}$

- Now substitute this into the other equation: $x={t}^{2}=(\frac{y}{2}{)}^{2}=\frac{{y}^{2}}{4}$

- So the cartesian equation is $x=\frac{{y}^{2}}{4}$

- It doesn't need to be in terms of $y$, this form is fine (or you could simplify to $4x={y}^{2}$)

## Method 2:

- This method is best for parametric equations containing trigonometric functions
- You'll need to add the two equations together - but it's best to first rearrange to a form where addition would create a trigonimetric identity
- Example: $x=2\mathrm{cos}\theta $ and $y=2\mathrm{sin}\theta $

- Looking at the right-hand sides of these equations, I can see that I'm adding $\mathrm{sin}$ and $\mathrm{cos}$

- I know that these functions can be added with the identity ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \equiv 1$

- So we need to get the equations to a form where this can be used

- First rearrange to $\frac{x}{2}=\mathrm{cos}\theta $ and $\frac{y}{2}=\mathrm{sin}\theta $

- Now square both sides to $\frac{{x}^{2}}{4}={\mathrm{cos}}^{2}\theta $ and $\frac{{y}^{2}}{4}={\mathrm{sin}}^{2}\theta $

- Now we can easily add the equations to get $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{4}={\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta $

- This simplifies to $\frac{{x}^{2}}{4}+\frac{{y}^{2}}{4}=1$

- Multiply both sides by 4: ${x}^{2}+{y}^{2}=4$

- Therefore, these equations form a circle with centre $(0,0)$ and radius 2

**g3** The parametric form of a circle:

- With method 2 above, you saw that the equations $x=2\mathrm{cos}\theta $ and $y=2\mathrm{sin}\theta $ formed a circle with centre $(0,0)$ and radius 2
- All parametric equations of the form $x=a+r\mathrm{cos}\theta $ and $y=b+r\mathrm{sin}\theta $ will form a circle with radius $r$ and centre $(a,b)$
- $a$ and $b$ can be any real number, so a circle with centre $(0,0)$ will have $a=0$ and $b=0$

**g4** Differentiation of parametric equations:

- Parametric equations can be differentiated with the chain rule (see C3)
- Your equation in terms of $x$ can be referred to as $\frac{\mathrm{d}x}{\mathrm{d}t}$ where $t$ is the parameter
- Your equation in terms of $y$ can be referred to as $\frac{\mathrm{d}y}{\mathrm{d}t}$ where $t$ is the parameter
- These can be used with the chain rule to calcluate $\frac{\mathrm{d}y}{\mathrm{d}x}$:

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}$ - The third fraction above is the reciprocal of $\frac{\mathrm{d}x}{\mathrm{d}t}$
- Therefore, we can derive that $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}$ (provided that $\frac{\mathrm{d}x}{\mathrm{d}t}$ $\ne 0$). It is helpful to memorise this - it's a useful shortcut

## Example:

- We'll find $\frac{\mathrm{d}y}{\mathrm{d}x}$ of $x=3{t}^{2}$ and $y=2{t}^{3}$
- Use C2-level differentiation to get $\frac{\mathrm{d}x}{\mathrm{d}t}$ $=3\times 2\times {t}^{2-1}=6t$

and $\frac{\mathrm{d}y}{\mathrm{d}t}$ $=2\times 3\times {t}^{3-1}=6{t}^{2}$ - Use the chain rule: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}t}\times \frac{\mathrm{d}t}{\mathrm{d}x}$ $=6{t}^{2}\times $ $\frac{1}{6t}=\frac{6{t}^{2}}{6t}$ $=t$
- Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x}$ $=t$

## Finding tangents and normals:

- Most parametric equation questions ask for the equation of the tangent or normal
- You can use the below from C1 do do this

- Use $y-{y}_{1}=m(x-{x}_{1})$ to find the tangent

- Use $y-{y}_{1}=-\frac{1}{m}(x-{x}_{1})$ to find the normal

# Calculus

**c1** The trapezium rule:

- ${\int}_{a}^{b}y\text{}\mathrm{d}x\approx \frac{1}{2}h$$\{({y}_{0}+{y}_{n})+2({y}_{1}+{y}_{2}+...+{y}_{n-1})\}$ where $h=$ $\frac{b-a}{n}$
- $h$ is the width of each strip - it's generally easier to memorise this fact rather than using $h=$ $\frac{b-a}{n}$ from the formula booklet
- Trapezium rule is covered in C2 if you need an explanation of using this formula (but the flashcards on this page still cover everything you need to know)
- Increasing the number of strips ($n$) will improve the accuracy

**c2** Integration with partial fractions:

- Some integrals can be solved by splitting them into partial fractions. An example is shown below

## Example:

- I'll be finding $\int \frac{3x+11}{{x}^{2}-x-6}$ $\mathrm{d}x$ in this example. I'll call the integral $I$
- If you use the method in section a5, you'll get $A=4$ and $B=-1$
- These can be used to write the integral in the form $I=$ $\int (\frac{4}{x-3}$ $-$ $\frac{1}{x+2})$ $\mathrm{d}x$
- We can now split the interval into two parts: $I=$ $\int \frac{4}{x-3}$ $\mathrm{d}x\text{}-$ $\int \frac{1}{x+2}$ $\mathrm{d}x$
- The first can be written as $\int 4(x-3{)}^{-1}\text{}\mathrm{d}x$. Use integration by substitution (with $u=x-3$) to integrate this to $4\mathrm{ln}|x-3|+c$
- The second can be written as $\int (x+2{)}^{-1}\text{}\mathrm{d}x$. Use integration by substitution (with $u=x+2$) to integrate this to $\mathrm{ln}|x+2|+c$
- The final step is to combine the two (remembering the $-$ sign from earlier). This gives us $I=4\mathrm{ln}|x-3|-\mathrm{ln}|x+2|+c$
- You could use the laws of logarithms to simplify this to $I=\mathrm{ln}$$|\frac{(x-3{)}^{4}}{x+2}|$ $+c$

**c3** Volumes of revolution:

- The
**volume of revolution**around an axis is the volume if a graph was rotated 360° about that axis. It is calculated with the below formulae - If rotated about the $x$-axis, $V={\int}_{a}^{b}\pi {y}^{2}\text{}\mathrm{d}x$ where $a$ and $b$ are points on the $x$-axis
- If rotated about the $y$-axis, $V={\int}_{a}^{b}\pi {x}^{2}\text{}\mathrm{d}y$ where $a$ and $b$ are points on the $y$-axis

**c4** Differential equations:

- A
**differential equation**is an equation which includes a derivative (e.g. $\frac{\mathrm{d}r}{\mathrm{d}t}$) in addition to other variables (e.g. $x$, $y$) - When formulating, remember that $\frac{\mathrm{d}a}{\mathrm{d}t}$ means
*the rate of change of $a$*

**c5** Solving first order differential equations:

- First, formulate the equation
- Then rearrange it so that all terms with $y$ are on the left of the $=$ and all terms with $x$ are on the right (or vice-versa)
- Integrate both sides (the left with respect to $y$ and the right with respect to $x$)
- Simplify and rearrange if required
- You only need one constant. For example, if your final form contains $+\text{}c+c$, you could simplify this to $+\text{}c$ because $c$ simply indicates any constant value

# Vectors

**v1** Notation:

**Vectors**represent a magnitude and direction (i.e. more than one dimension)**Scalars**represent only a magnitude (i.e. one direction only)- Vectors are written as $\mathbf{v}$ (bold) or $\overrightarrow{v}$ with an arrow above
- For example, to get from point $A$ to point $B$, you would use the vector $\overrightarrow{AB}$
- In a diagram, an arrow is often used - the length represents the magnitude and it points in the direction of the vector
- The magnitude of a vector $\mathbf{v}$ is written as $|\mathbf{v}|$ (modulus $\mathbf{v}$). It is calculated with Pythagoras' theorem; $|\mathbf{v}|=\sqrt{{x}^{2}+{y}^{2}\phantom{\rule{.1667em}{0ex}}[+{z}^{2}]}$

## Unit vectors:

- A
**unit vector**is a vector with a magnitude of 1 - $\mathbf{i}$ represents the $x$ direction, and $\mathbf{j}$ the $y$ direction
- For example, $5\mathbf{i}+4\mathbf{j}$ represents the vector $\left(\begin{array}{c}5\\ 4\end{array}\right)$, 5 right and 4 up. This doesn't have to start at the origin, it can be a translation from one point to another
- For 3D vectors, $\mathbf{k}$ is used to represent the unit vector for the third dimension

## Position vectors:

- A
**position vector**is the distance and direction of a point from the origin, $O$

## Equal vectors:

- Two vectors are equal if they have the same magnitude and direction

**v2** Adding and multiplying:

- If you multiply a vector by a scalar, its magnitude (length) is multiplied by the scalar. The direction remains the same
- If you add two vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$, you'll get a single vector which is $\overrightarrow{AC}$

**v3** Scalar product:

- If you multiply two vectors together to get a scalar, you have the
**scalar product**(also known as*dot product*) - With the vectors $\mathbf{a}=\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\end{array}\right)$, the scalar product is $\mathbf{a}\cdot \mathbf{b}={a}_{1}{b}_{1}+{a}_{2}{b}_{2}$
- This also works in 3D; with $\mathbf{a}=\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right)$, the scalar product is $\mathbf{a}\cdot \mathbf{b}={a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}$
- The angle between two vectors can be calculated by rearranging $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}|\mathrm{cos}\theta $

Note: the formula booklet has a similar formula in the*Vector product*section, but containing $\mathrm{sin}\theta $ instead of $\mathrm{cos}\theta $.**Do not copy it, it is for the FP3 module** - If two non-zero vectors are perpendicular, $\mathbf{a}\cdot \mathbf{b}=0$

**v4** Distances:

- The distance between the points $({x}_{1},{y}_{1},{z}_{1})$ and $({x}_{2},{y}_{2},{z}_{2})$ is $\sqrt{({x}_{1}-{x}_{2}{)}^{2}+({y}_{1}-{y}_{2}{)}^{2}+({z}_{1}-{z}_{2}{)}^{2}}$

**v5** The vector equation of a line:

- First, calculate the co-ordinates of two points on the line if you don't already have them
- Then substitute them into the equation $r=\overrightarrow{OA}+\lambda \overrightarrow{AB}$
- So if your two points are $A(a,b)$ and $B(c,d)$, the equation is $r=\left(\begin{array}{c}a\\ b\end{array}\right)+\lambda \left(\begin{array}{c}c-a\\ d-b\end{array}\right)$
- $\lambda $ can be any real number. Each different value of it will produce a different set of $(x,y)$ co-ordinates on the line

## Converting to cartesian form:

- The vector equation $r=\left(\begin{array}{c}a\\ b\\ [c]\end{array}\right)+\lambda \left(\begin{array}{c}d\\ e\\ [f]\end{array}\right)$ in cartesian form is $\frac{x-a}{d}$ $=$ $\frac{y-b}{e}$ [$=$ $\frac{z-c}{f}$] (ignore everything in the square brackets if you're only dealing with a 2D vector)

**v6** The vector equation of a plane:

- You can write the vector equation of a plane given three vectors:

-**1**a position vector from the origin to a point $\mathbf{A}$ on the plane

-**2**a vector from $\mathbf{A}$ to another point $\mathbf{B}$ on the plane

-**3**a vector from $\mathbf{A}$ to another point $\mathbf{C}$ on the plane which is**not**parallel to the vector in**2** - With the vectors defined above, the vector equation of the plane containing $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ is $r=\overrightarrow{OA}+\lambda \overrightarrow{AB}+\mu \overrightarrow{AC}$

## Converting a vector equation of a plane to a cartesian equation:

- To do this, use ${n}_{1}x+{n}_{2}y+{n}_{3}z+d=0$ where $d=-\mathbf{a}\cdot \mathbf{n}$

- $\mathbf{a}$ is the position vector of a point on the plane

- $\mathbf{n}$ is a vector**perpendicular**to the plane

- ${n}_{1}$, ${n}_{2}$ and ${n}_{3}$ are the elements in the vector $\mathbf{n}$ - Instead of calculating $d$ with the above formula, you could also substitute in the values of $x$, $y$ and $z$ from a known point on the plane and solving for $d$
- This can also be written as $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ where

- $\mathbf{r}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$

- $\mathbf{a}$ and $\mathbf{n}$ are as above

**v7** Perpendicular vectors:

- A vector which is perpendicular to a plane is perpendicular to any line in the plane
- So, if a vector is perpendicular to two non-parallel lines in a plane, it is perpendicular to the plane

**v8** The angle between two planes:

- The angle between two planes is the same as the angle between their normals

**v9** The intersection of a line and a plane:

- If you have a line $\mathbf{r}=\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\right)$ $+$ $\lambda \left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right)$ and a plane ${n}_{1}x+{n}_{2}y+{n}_{3}z=d$, you can use the following method to find the points where they intersect:
- First, set $r=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ to create the equation $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=$ $\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\right)$ $+$ $\lambda \left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right)$
- Now create three simultaneous equations from this:

$x={a}_{1}+{b}_{1}\lambda $

$y={a}_{2}+{b}_{2}\lambda $

$z={a}_{3}+{b}_{3}\lambda $ - Now substitute these values into the cartesian equation of the plane, i.e. ${n}_{1}({a}_{1}+{b}_{1}\lambda )$ $+$ ${n}_{2}({a}_{2}+{b}_{2}\lambda )$ $+$ ${n}_{3}({a}_{3}+{b}_{3}\lambda )=d$
- If you expand the brackets, simplify and rearrange in terms of $\lambda $, you'll get a numerical value for $\lambda $
- Now substitute this value of $\lambda $ into the original vector equation of the line. This will give you the $x$, $y$ and $z$ coordinates of the point of intersection
- You can check your answer by substituting these values into the cartesian equation of the plane